Find the angular speed of earth so that a body lying at `30^(@)` latitude may become weightless.

Answer

To find the angular speed of the Earth so that a body lying at 30 degrees latitude may become weightless, we can follow these steps:

### Step 1: Understand the concept of weightlessness
A body is said to be weightless when the effective gravitational force acting on it becomes zero. The effective weight (W) of the body can be expressed as:
\[ W = m \cdot g_{\text{effective}} \]
where \( g_{\text{effective}} \) is the effective gravitational acceleration.

### Step 2: Express the effective gravitational acceleration
The effective gravitational acceleration \( g_{\text{effective}} \) can be expressed as:
\[ g_{\text{effective}} = g - R \cdot \omega^2 \cdot \cos^2(\lambda) \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (approximately \( 6.4 \times 10^6 \, \text{m} \)),
- \( \omega \) is the angular speed of the Earth,
- \( \lambda \) is the latitude (in this case, \( 30^\circ \)).

### Step 3: Set the effective gravitational acceleration to zero
For the body to be weightless:
\[ g - R \cdot \omega^2 \cdot \cos^2(30^\circ) = 0 \]

### Step 4: Calculate \( \cos(30^\circ) \)
Using the trigonometric identity:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute \( \cos(30^\circ) \) into the equation
Substituting \( \cos(30^\circ) \) into the equation gives:
\[ g - R \cdot \omega^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = 0 \]
This simplifies to:
\[ g - R \cdot \omega^2 \cdot \frac{3}{4} = 0 \]

### Step 6: Rearrange the equation to solve for \( \omega^2 \)
Rearranging the equation gives:
\[ R \cdot \omega^2 = \frac{4g}{3} \]
Thus,
\[ \omega^2 = \frac{4g}{3R} \]

### Step 7: Take the square root to find \( \omega \)
Taking the square root of both sides gives:
\[ \omega = \sqrt{\frac{4g}{3R}} \]

### Step 8: Substitute the known values
Substituting \( g \approx 9.81 \, \text{m/s}^2 \) and \( R \approx 6.4 \times 10^6 \, \text{m} \):
\[ \omega = \sqrt{\frac{4 \times 9.81}{3 \times 6.4 \times 10^6}} \]

### Step 9: Calculate \( \omega \)
Calculating the above expression gives:
\[ \omega \approx \sqrt{\frac{39.24}{19200000}} \]
\[ \omega \approx \sqrt{2.046 \times 10^{-6}} \]
\[ \omega \approx 0.00143 \, \text{rad/s} \]

Thus, the angular speed of the Earth for a body lying at 30 degrees latitude to become weightless is approximately \( 0.00143 \, \text{rad/s} \).